Question

3. The base AB of two equilateral triangles ABC and ABD with side
2a, lies along the x-axis such that the mid point of AB is at the
origin. Find the coordinates of two vertices C and D of the
triangles​

Answer 1

Answer:

Since the mid-point of AB is at the origin O and AB=2a

∴    OA=OB=a.

Thus, the coordinates of A and B are (a,0) and (−a,0) respectively.

Since triangles ABC. and ABC' are equilateral. Therefore, their third vertices C and C'

lie on the perpendicular bisector of base AB. Clearly, YOY is the perpendicular bisector

of AB. Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.

In △AOC,wehave

OA2+OC2=AC2

⇒a2+OC2=(2a)2

⇒OC2=4a2−a2

⇒OC2=3a2

OC=3a

​

Similarly, by applying Pythagoras theorem in △AOC; we have OC′=3a

​

Thus, the coordinates of C and C' are (0,3

​a)and(0,−3

​a)respectively.

The base AB of the two equilateral triangles ABC and ABC' with side 2a lies along the X-axis such that the mid-point of AB is at the origin. ... Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.

Step-by-step explanation:

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Answer 2

Since the mid-point of AB is at the origin O and AB=2a

∴    OA=OB=a.

Thus, the coordinates of A and B are (a,0) and (−a,0) respectively.

Since triangles ABC. and ABC' are equilateral. Therefore, their third vertices C and C'

lie on the perpendicular bisector of base AB. Clearly, YOY is the perpendicular bisector

of AB. Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.

In △AOC,wehave

OA2+OC2=AC2

⇒a2+OC2=(2a)2

⇒OC2=4a2−a2

⇒OC2=3a2

OC=3a

Similarly, by applying Pythagoras theorem in △AOC; we have OC′=3a

Thus, the coordinates of C and C' are (0,3a)and(0,−3a)respectively.